Let $R$ be the ring $R[X,Y]/(X^2+Y^2−1)$. The space of $\mathbb{R}$-rational points of the affine scheme associated to $R$ is the topological circle $S^1$.

An algebraic vector bundle over $R$ is an $R$-algebra $A$ with certain properties or equivalently a finitely generated projective $R$-module $A$.

My first question is: What is the explicit projective $R$-module $A$ corresponding to the topological Moebius bundle over $S^1$? (By 'corresponding' I mean in particular that the $\mathbb{R}$-rational points should induce the topological Moebius bundle and that it has rank one.)

My second question is motivated by the fact that topologically there are only two non-isomorphic rank one bundles over $S^1$: the trivial bundle and the Moebius bundle. What is the analogous algebraic situation? Are there more than two non-isomorphic and rank one projective modules over $R$?

Thank you!

(I apologize that I first asked this question on math.stackexchange, I deleted it there.)

4 Answers
4

The internal note "Le ruban de Moebius comme représentation d'un idéal non principal" (Moebius band, as a non-principal ideal), by Daniel Ferrand, contains more or less the material you want. Daniel is retired and doesn't have a website, but, thank you Google!,
the note is downloadable from http://www.math.unibas.ch/~giordano/Moebius.pdf

The non-principal ideal you want is the ideal $(1+x,y)$ in the ring $R[x,y]/(x^2+y^2-1)$.

I am a little confused: Tom Goodwillie's answer describes the Moebius $R$-module $A$ as the image of a morphism $R^2\to R^2$ given by a 2x2-matrix, if I am not missing something. You give a non-principal ideal $I$ of $R$. This corresponds to $A$ by the ideal-class-group correspondence mentioned in the answers. But this does not mean that $A=R/I$, right? If this is not true, I am totally confused: $R/I$ has less prime ideals than $R$ and hence can't be a bundle. Thank you.
–
roger123Nov 9 '10 at 20:48

2

The ideal $I$ is isomorphic, as an $R$-module, to the submodule $A$ of $R^2$ defined as the image of Tom's projector. (Eg, by mapping $a(1+x)+by\in I$ to the element $2 p(a,b)$ in $R^2$.)
–
ACLNov 9 '10 at 21:40

As to your second question, the "analogous algebraic situation" is the following result.

Theorem: Let $k$ be a field of characteristic different from $2$ in which $-1$ is not a square (e.g. $k = \mathbb{R}$!). Then the Dedekind domain $R_{\circ} = k[x,y]/(x^2+y^2-1)$ is not a PID, but rather has ideal class group of order $2$. In other words, there is up to isomorphism exactly one nontrivial algebraic line bundle over the corresponding affine curve.

See http://math.uga.edu/~pete/ellipticded.pdf for some treatment of this result, including deduction of it from a more general result of Rosen. That this is the algebraic analogue of the fact that the Mobius band is the unique nontrivial topological line bundle over the unit circle is actually mentioned there. It is also related to the nonunique factorization

$(1+\sin \theta)(1-\sin \theta) = (\cos \theta)(\cos \theta)$.

As to the explicit construction of the corresponding rank one module: you are looking for a nonprincipal ideal in the Dedekind domain $R$. I think one should be able to use the non-unique factorization into irreducible elements above (here of course we identify $\cos \theta$ with $x$ and $\sin \theta$ with $y$) to write down an explicit nonprincipal ideal in $R$. Try it, and if you don't succeed and don't get another, better answer in the meantime, let me know.

Here is the explicit construction. Let $T$ be the ring of $4\pi$-periodic continuous functions on $\mathbb{R}$ and $S$ be the ring of $2\pi$-periodic continuous functions on $\mathbb{R}$. Then $T$ is a free $S$-module of rank 2. Mobious bundle corresponds to a submodule $T_1$ of $T$ consisting of all $2\pi$-antiperiodic functions:

Now if you consider $S$ as a structure sheaf on $\mathbb{R}$ then $T_1$ becomes a sheaf of $S$-modules, which is locally free of rank 1 but is not trivial globally. In fact, continuity implies that any section of $T_1$ must have a zero somewhere. This construction also shows how it is possible that a projective module is not free itself.